---- > [!definition] Definition. ([[coset]]) > If $H$ is a [[subgroup]] of a [[group]] $G$, and $a \in G$, the subset $aH := \{ ah : h \in H \}$ is called a **left $H$-coset** of $H$. Likewise, the subset $Ha := \{ ha : h \in H \}$ > is called a **right $H$-coset** of $H$. We call $a$ the **representative** of these [[coset|cosets]]. > \ > The [[subgroup]] $H$ is a particular **left/right $H$-coset** because $1H=H=H1$. > \ > **Notation.** In light of [[characterization of cosets]], the set of [[coset]]s of $H$ in $G$ by $G/H$ is denoted $G / H$. Its [[cardinality]] is the [[index of a subgroup|index]] of $H$ in $G$. > \ > **Remark.** In a [[Cayley diagram]], the **left coset** $aH$ can be found as follows: *start from node $a$ and follow **all** paths in $H$*. ![[CleanShot 2023-09-10 at 13.36.12@2x 1.jpg]] > [!intuition] > The verbal idea for obtaining the coset $aH$ or $Ha$ is to "fix $a$, then multiply through all elements of $H$ by $a$... now look at the elements of $G$ obtained in this way." > [!basicexample] Example. (Cosets of $S_{3}$) Denote by $G:=S_{3}$ the [[symmetric group]] on $3$ letters $\{ e,\tau, \tau^{2}, \sigma, \sigma \tau, \sigma \tau^{2} \}$. It has proper, nontrival [[subgroup]]s $\langle \tau \rangle= \{ e, \tau, \tau^{2}\}, \langle \sigma \rangle = \{ e, \sigma \}, \langle \sigma \tau \rangle=\{ e, \sigma \tau \}, \langle \sigma \tau^{2} \rangle=\{ e, \sigma \tau^{2} \}$, along with the obvious [[subgroup]]s $G$ and $\{ e \}$. *Let's find all left and right cosets of $S_{3}$.* # Left and Right Cosets of $G$ Let $a,b \in G$. Start with $G$. $aG$ 'cycles through the elements of $G$, as does $Gb$. So $aG=G=Gb$ for any $a,b \in G$. # Left and Right Cosets of $\{ e \}$ The left *and* right cosets of $\{ e \}$ will be the singletons $\{ e \}$, $\{ \tau \}$, $\tau^{2}$, $\{ \sigma \}$, $\{ \sigma \tau \}$, $\{ \sigma \tau^{2} \}$. **Remark.** In each of the above two 'extremal cases', the left cosets equaled the right cosets. Below we see that this is not true in general. Let us relabel the remaining [[subgroup]]s as follows: - $K:=\{ e,\tau, \tau^{2} \}$; - $H_{1}:= \{ e, \sigma \}$; - $H_{2}:=\{ e, \sigma \tau \}$; - $H_{3}:=\{ e, \sigma \tau, \sigma \tau^{2} \}$. # Left Cosets of $K$ - $\textcolor{Thistle}{eK = K}$; - $\textcolor{Thistle}{\tau K = \{ \tau e, \tau \tau, \tau \tau \tau \} = K}$; - $\textcolor{Thistle}{\tau^{2}K = K}$ (same idea^); - $\textcolor{Skyblue}{\sigma K = \{ \sigma, \sigma \tau, \sigma \tau^{2} \}}$; - $\textcolor{Skyblue}{(\sigma \tau)K = \sigma(\tau K) = \sigma K = \{ \sigma, \sigma \tau, \sigma \tau^{2} \}}$; - $\textcolor{Skyblue}{(\sigma \tau^{2})K = \sigma (\tau^{2}K) = \sigma K = \{ \sigma, \sigma \tau, \sigma \tau^{2} \}}$. **Remark.** The left cosets above, color-coded by equality (two distinct), are disjoint and union to $G$. # Right Cosets of $K$ - $\textcolor{Thistle}{Ke=K}$; - $\textcolor{Thistle}{K\tau=\{ e \tau, \tau \tau, \tau \tau \tau \}=K}$; - $\textcolor{Thistle}{K \tau^{2} = K}$ (same idea^) - $\textcolor{Skyblue}{K \sigma=\{e\sigma, \tau \sigma, \tau^{2} \sigma \}=\{ \sigma, \sigma \tau^{2}, \sigma \tau \}}$; - $\textcolor{Skyblue}{K(\sigma \tau)=(K\sigma)\tau=\{ \sigma, \sigma \tau^{2}, \sigma \tau \}\tau=\{ \sigma \tau, \sigma, \sigma \tau^{2} \}}$; - $\textcolor{Skyblue}{K(\sigma \tau^{2})=(K\sigma)\tau^{2}=\{ \sigma, \sigma \tau^{2}, \sigma \tau \}\tau^{2}=\{ \sigma \tau^{2},\sigma \tau, \sigma \}}$. **Remark.** Similarly, the right cosets above, color-coded by equality (two distinct), are disjoint and union to $G$. **Remark.** In each of the above 'extremal cases' and the case of $K$, the left cosets equaled the right cosets. Below we see that this is not true in general. The $H_{i}$ are similar enough (think of just as reflections over different axes) that we will only treat one case. So set $H:=H_{1}=\{ e,\sigma \}$. # Left Cosets of $H$ - $\textcolor{Goldenrod}{eH=\{ e, \sigma \}=H}$; - $\textcolor{Thistle}{\tau H= \{ \tau e, \tau \sigma \}=\{ \tau, \sigma \tau^{2} \}}$; - $\textcolor{Skyblue}{\tau^{2} H=\{ \tau^{2}e, \tau^{2} \sigma \}=\{ \tau^{2}, \sigma \tau \}}$; - $\textcolor{Goldenrod}{\sigma H= \{ e , \sigma^{2}\}=H}$; - $\textcolor{Skyblue}{(\sigma \tau )H=\sigma(\tau H)=\{ \sigma \tau, \tau^{2} \}}$; - $\textcolor{Thistle}{(\sigma \tau^{2} )H=\sigma (\tau^{2} H)= \{ \sigma \tau^{2}, \tau \}}$. **Remark.** The left cosets above, color-coded by equality (three distinct), are disjoint and union to $G$. They're shaded accordingly in the [[Cayley diagram]] below: ![[CleanShot 2023-09-10 at 11.35.25.jpg]] From this picture it is clear that the [[index of a subgroup|index]] of $H$ in $G$ is $3$. # Right Cosets of $H$ - $\textcolor{Goldenrod}{He=\{ e, \sigma \}=H}$; - $\textcolor{Teal}{H\tau=\{ \tau, \sigma \tau \}}$; - $\textcolor{Coral}{H \tau^{2} = \{ \tau^{2}, \sigma \tau^{2} \}}$; - $\textcolor{Goldenrod}{H \sigma = \{ e, \sigma \}=H}$; - $\textcolor{Teal}{H(\sigma \tau)=H\tau= \{ \tau, \sigma \tau \}}$; - $\textcolor{Coral}{H(\sigma \tau^{2})=H \tau^{2}= \{ \tau^{2}, \sigma \tau ^{2} \}}$. **Remark.** Similarly, the right cosets above, color-coded by equality (three distinct), are disjoint and union to $G$. # Big Idea We observe in all cases above that any two left cosets of a subset of $S_{3}$ are either disjoint or identical. Moreover, [[cardinality|cardinalities]] always match $|H|$. This turns out to be generally true, and is key to proving [[Lagrange's Theorem]]. # An Example with the [[dihedral group]] ![[CleanShot 2023-09-10 at [email protected]]] ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```